Question

A body of mass 2 kg is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in 4s is $\frac{1}{3}\alpha^2\sqrt{P}m$. The value of $\alpha$ will be ____.

Answer 1

4

Answer 2

For a constant power source, the velocity of the body is given by $v = \sqrt{\frac{2Pt}{m}}$. The displacement is $s = \int_0^t v dt = \int_0^t \sqrt{\frac{2Pt'}{m}} dt' = \frac{2}{3}\sqrt{\frac{2}{m}} P^{1/2} t^{3/2}$. Given $m=2$ kg, $t=4$ s, and $s = \frac{1}{3}\alpha^2\sqrt{P}$ m. Substituting values: $s = \frac{2}{3}\sqrt{\frac{2}{2}} P^{1/2} (4)^{3/2} = \frac{2}{3} \times 1 \times \sqrt{P} \times 8 = \frac{16}{3}\sqrt{P}$. Equating this to the given displacement: $\frac{16}{3}\sqrt{P} = \frac{1}{3}\alpha^2\sqrt{P}$. This gives $\alpha^2 = 16$, so $\alpha = 4$.